Как найти интеграл от sin2x

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int sin2x dx

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    Источник

    The integral of sin 2x and the integral of sin2x have different values. To find the integral of sin2x, we use the cos 2x formula and the substitution method whereas we use just the substitution method to find the integral of sin 2x.

    Let us identify the difference between the integral of sin 2x and the integral of sin2x by finding their values using appropriate methods and also we will solve some problems related to these integrals.

    1. What is the Integral of sin 2x dx?
    2. Definite Integral of sin 2x
    3. What is the Integral of sin^2x dx?
    4. Definite Integral of sin^2x
    5. FAQs on Integral of sin 2x and sin^2x

    What is the Integral of Sin 2x dx?

    The integral of sin 2x is denoted by ∫ sin 2x dx and its value is -(cos 2x) / 2 + C, where ‘C’ is the integration constant. For proving this, we use the integration by substitution method. For this, we assume that 2x = u. Then 2 dx = du (or) dx = du/2. Substituting these values in the integral ∫ sin 2x dx,

    ∫ sin 2x dx = ∫ sin u (du/2)

    = (1/2) ∫ sin u du

    We know that the integral of sin x is -cos x + C. So,

    = (1/2) (-cos u) + C

    Substituting u = 2x back here,

    ∫ sin 2x dx = -(cos 2x) / 2 + C

    This is the integral of sin 2x formula.

    Integral of sin 2x dx formula is -cos 2x over 2 all plus C.

    Definite Integral of Sin 2x

    A definite integral is an indefinite integral with some lower and upper bounds. By the fundamental theorem of Calculus, to evaluate a definite integral, we substitute the upper bound and the lower bound in the value of the indefinite integral and then subtract them in the same order. While evaluating a definite integral, we can ignore the integration constant. Let us calculate some definite integrals of integral sin 2x dx here.

    Integral of Sin 2x From 0 to pi/2

    ∫(_0^{pi/2}) sin 2x dx = (-1/2) cos (2x) (left. right|_0^{pi/2})

    = (-1/2) [cos 2(π/2) — cos 2(0)]

    = (-1/2) (-1 — 1)

    = (-1/2) (-2)

    = 1

    Therefore, the integral of sin 2x from 0 to pi/2 is 1.

    Integral of Sin 2x From 0 to pi

    ∫(_0^{pi}) sin 2x dx = (-1/2) cos (2x) (left. right|_0^{pi})

    = (-1/2) [cos 2(π) — cos 2(0)]

    = (-1/2) (1 — 1)

    = 0

    Therefore, the integral of sin 2x from 0 to pi is 0.

    What is the Integral of Sin^2x dx?

    The integral of sin2x is denoted by ∫ sin2x dx and its value is (x/2) — (sin 2x)/4 + C. We can prove this in the following two methods.

    • By using the cos 2x formula
    • By using the integration by parts

    The integral of cos square x dx is x over 2 all plus sin 2x over 4 all plus C.

    Method 1: Integral of Sin^2x Using Double Angle Formula of Cos

    To find the integral of sin2x, we use the double angle formula of cos. One of the cos 2x formulas is cos 2x = 1 — 2 sin2x. By solving this for sin2x, we get sin2x = (1 — cos 2x) / 2. We use this to find ∫ sin2x dx. Then we get

    ∫ sin2x dx = ∫ (1 — cos 2x) / 2 dx

    = (1/2) ∫ (1 — cos 2x) dx

    = (1/2) ∫ 1 dx — (1/2) ∫ cos 2x dx

    We know that ∫ cos 2x dx = (sin 2x)/2 + C. So

    ∫ sin2x dx = (1/2) x — (1/2) (sin 2x)/2 + C (or)

    ∫ sin2x dx = x/2 — (sin 2x)/4 + C

    This is the integral of sin^2 x formula. Let us prove the same formula in another method.

    Method 2: Integral of Sin^2x Using Integration by Parts

    We know that we can write sin2x as sin x · sin x. To find the integral of a product, we can use the integration by parts.

    ∫ sin2x dx = ∫ sin x · sin x dx = ∫ u dv

    Here, u = sin x and dv = sin x dx.

    Then du = cos x dx and v = -cos x.

    By integration by parts formula,

    ∫ u dv = uv — ∫ v du

    ∫ sin x · sin x dx = (sin x) (-cos x) — ∫ (-cos x)(cos x) dx

    ∫ sin2x dx = (-1/2) (2 sin x cos x) + ∫ cos2x dx

    By the double angle formula of sin, 2 sin x cos x = sin 2x and by a trigonometric identity, cos2x = 1 — sin2x. So

    ∫ sin2x dx = (-1/2) sin 2x + ∫ (1 — sin2x) dx

    ∫ sin2x dx = (-1/2) sin 2x + ∫ 1 dx — ∫ sin2x dx

    ∫ sin2x dx + ∫ sin2x dx = (-1/2) sin 2x + x + C₁

    2 ∫ sin2x dx = x — (1/2) sin 2x + C₁

    ∫ sin2x dx = x/2 — (sin 2x)/4 + C (where C = C₁/2)

    Hence proved.

    Definite Integral of Sin^2x

    To evaluate the definite integral of sin2x, we just substitute the upper and lower bounds in the value of the integral of sin2x and subtract the resultant values. Let us evaluate some definite integrals of integral sin2x dx here.

    Integral of Sin^2x From 0 to 2pi

    ∫(_0^{2pi}) sin2x dx = [x/2 — (sin 2x)/4] (left. right|_0^{2pi})

    = [2π/2 — (sin 4π)/4] — [0 — (sin 0)/4]

    = π — 0/4

    = π

    Therefore, the integral of sin2x from 0 to 2π is π.

    Integral of Sin^2x From 0 to pi

    ∫(_0^{pi}) sin2x dx = [x/2 — (sin 2x)/4] (left. right|_0^{pi})

    = [π/2 — (sin 2π)/4] — [0 — (sin 0)/4]

    = π/2 — 0/4

    = π/2

    Therefore, the integral of sin2x from 0 to π is π/2.

    Important Notes Related to Integral of Sin 2x and Integral of Sin2x:

    • ∫ sin 2x dx = -(cos 2x)/2 + C
    • ∫ sin2x dx = x/2 — (sin 2x)/4 + C

    Topics Related to Integral of Sin2x and Integral of Sin 2x:

    • Indefinite Integral
    • Integral Calculator
    • Indefinite Integral Calculator
    • Definite Integral Calculator
    • Applications of Integrals

    FAQs on Integral of Sin 2x and Sin^2x

    What is the Integral of Sin 2x dx?

    The integral of sin 2x dx is written as ∫ sin 2x dx and ∫ sin 2x dx = -(cos 2x)/2 + C, where C is the integration constant.

    What is the Integral of Sin^2x dx?

    The integral of sin^2x dx is written as ∫ sin2x dx and ∫ sin2dx = x/2 — (sin 2x)/4 + C. Here, C is the integration constant.

    How to Find the Definite Integral of Sin 2x from 0 to Pi/4?

    We know that ∫ sin 2x dx = -(cos 2x)/2. Substituting the limits 0 and π/4, we get (-1/2) cos 2(π/4) — (-1/2) cos 2(0) = (1/2) 0 + (1/2) (1) = 1/2.

    What is the Integral of Sin^3x dx?

    ∫ sin3x dx = ∫ sin2x sin x dx = ∫ (1 — cos2x) sin x dx. Let us substitute cos x = u. Then -sin x dx = du. Then the above integral becomes, ∫ (1 — u2) (- du) = -u + u3/3 + C. Substituting u = sin x back here, ∫ sin3x dx = -cos x + cos3x/3 + C.

    What is the Integral of Sin 3x dx?

    To find the ∫ sin 3x dx, let that 3x = u. Then 3 dx = du. From this, dx = du/3. Then the above integral becomes, ∫ sin u (1/3) du = (1/3) (-cos u) + C = (-1/3) cos (3x) + C.

    How to Find the Definite Integral of sin^2x from 0 to Pi/4?

    We know that ∫ sin2dx = x/2 — (sin 2x)/4 + C. Substituting the limits here, we get [π/8 — (sin π/2)/4] — [0 — (sin 0)/4] = π/8 — 1/4.

    Is the Integral sin 2x dx Same as Integral sin^2x dx?

    No, the values of these two integrals are NOT same. We have

    • ∫ sin2dx = x/2 — (sin 2x)/4 + C
    • ∫ sin 2x dx = (-cos 2x)/2 + C
    Источник

    Integration of sin2x means finding the integral of the function sin2x. Integral of sin2x can be written as ∫ sin2x dx. Here, we need to find the indefinite integral of sin2x. So, the integration of sin2x results in a new function with arbitrary constant C. Let’s learn the formula for the integration of sin2x along with the derivation.

    The formula for the integral of sin 2x dx is given by:

    ∫ sin2x dx = -(½) cos2x + C

    Learn: Integral

    Integration of Sin 2x Derivation

    We can derive the formula of integral of sin2x using the method of integration by substitution.

    Consider ∫sin2x dx

    Let u = 2x such that du = 2dx ⇒ dx = (½)du.

    Substituting the above equations in ∫sin2x dx, we get;

    ∫ sin2x dx = ∫ sin u (½) du

    = (½) ∫ sin u du

    = (½) (-cos u) + C

    = -(½) cos2x + C

    Therefore, ∫ sin2x dx = -(½) cos2x + C

    Solved Examples

    Let’s have a look at the examples given below to understand how to apply the formula of integration of sin 2x.

    Example 1: Integration of Sin2x dx from 0 to pi/2

    Integration of sin2x dx from 0 to pi/2 can be written as:

    (begin{array}{l}int_{0}^{frac{pi}{2}}sin2x dxend{array} )

    This is the definite integral of sin2x with lower limit 0 and upper limit π/2.

    We know that, ∫ sin2x dx = -(½) cos2x + C

    So,

    (begin{array}{l}int_{0}^{frac{pi}{2}}sin2x dx=left [ -(frac{1}{2})cos2x right ]_{0}^{frac{pi}{2}}end{array} )

    Now, apply the limits to the new function.

    (begin{array}{l}=-(frac{1}{2})left [ cos2(frac{pi}{2})-cos2(0) right ]end{array} )

    = -(½) [cos π – cos 0]

    -(½) [-1 – 1]

    = -(½)(-2)

    = 1

    Therefore, integration of sin 2x from o to pi/2 is equal to 1.

    Example 2: Integration of Sin(2x+1)

    Integration of sin(2x+1) can be written as: ∫ sin(2x + 1)dx

    We know that, ∫ sin2x dx = -(½) cos2x + C

    So, ∫ sin(2x + 1) dx = -(½) cos(2x+1) + C

    Example 3: Integration of Sin2x/1+cosx

    Integration of Sin2x/1+cosx

    = ∫ (sin2x)/(1 + cos x) dx

    Using the sin2x formula, i.e. sin2x = 2 sinx cosx

    = ∫ (2 sinx cosx)/(1 + cosx) dx

    = 2 ∫[cosx/(1 + cos x)] sinx dx

    Let u = cos x

    du = -sinx dx

    Susbtituting these values, we get;

    = 2 ∫[u/(1 + u)] (-du)

    = -2 ∫ (u + 1 – 1)/(u + 1) du

    = -2 ∫ (u + 1)/(u + 1) du + 2 ∫ 1/(u + 1) du

    = -2 ∫ 1 du + 2 log|u + 1|

    = -2u + 2log|u + 1| + C

    = -2 cosx + 2log|cosx + 1| + C

    Источник

    ∫sin2x dx

    To integrate sin2x, also written as ∫sin2x dx, and sin 2x, we usually use a u
    substitution to build a new integration in terms of u.

    u=2x

    Let u=2x.

    du/dx

    Then du/dx = 2

    dx

    We rearrange to get an expression for dx in terms of u.

    New integration in terms of u.

    As you can see, we now have a new integration in terms of u, which means the
    same thing. We get this by
    replacing 2x with u, and replacing dx with ½ du.

    moving ½ outside of the integral sign.

    We move the ½ outside of the integral as it is simply a multiplier.

    integral of sinu

    We now have a simple integration with sinu that we can solve easily.

    Reintroduce ½ constant.

    We remember the ½ outside the integral sign and reintroduce it here.

    -½cos2x + C

    Hence, this is the final answer -½cos2x + C, where C is the integration
    constant.

    Источник

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    Интеграл от синуса

    Интеграл от синуса по таблице интегрирования равен: $$ int sin x dx = — cos x + C $$

    Словами это читается так: интеграл от синуса равен сумме отрицательного косинуса и произвольной постоянной.

    Пример 1
    Найти интеграл от синус 2x: $$ int sin 2x dx $$
    Решение

    Напрямую интеграл взять не получится, так как аргумент синуса и знака дифференциала отличаются. Выполняем подведение под дифференциал $ 2x $ и добавляем перед интегралом дробь $ frac{1}{2} $:

    $$ int sin 2x dx = frac{1}{2} int sin 2x d(2x) = -frac{1}{2} cos 2x + C $$

    Если не получается решить свою задачу, то присылайте её к нам. Мы предоставим подробное решение онлайн. Вы сможете ознакомиться с ходом вычисления и почерпнуть информацию. Это поможет своевременно получить зачёт у преподавателя!
    Ответ
    $$ int sin 2x dx = -frac{1}{2} cos 2x + C $$
    Пример 2
    Найти интеграл от синуса в квадрате: $$ int sin^2 x dx $$
    Решение

    В данном случае необходимо воспользоваться одной из тригонометрических формул. Конкретно формулой понижения степени синуса: $$ sin^2 x = frac{1-cos 2x}{2} $$

    Заменяем выражение под интегралом:

    $$ int sin^2 x dx = int frac{1-cos 2x}{2} dx = frac{1}{2} int (1-cos 2x) dx = $$

    $$ = frac{1}{2} int 1dx — frac{1}{2} int cos 2x dx = frac{1}{2}x — frac{1}{2}cdotfrac{1}{2}int cos 2x d(2x) = $$

    $$ = frac{1}{2}x — frac{1}{4}sin 2x + C $$
    Ответ
    $$ int sin^2 x dx = frac{1}{2}x — frac{1}{4}sin 2x + C $$
    Пример 3
    Найти интеграл от синуса в кубе: $$ int sin^3 x dx $$
    Решение

    Здесь нужно вспомнить свойство степеней и учесть: $$ sin^3 x = sin x cdot sin^2 x $$

    Подставляем, полученное выражение в интеграл и заносим $ sin x $ под знак дифференциала:

    $$ int sin^3 x dx = int sin x sin^2 x dx = — int sin^2 x d(cos x) = $$

    Далее используем свойство $ sin^2 x = 1 — cos^2 x $:

    $$ = -int (1-cos^2 x) d(cos x) = -int d(cos x) + int cos^2 x d(cos x) = $$

    $$ = — cos x + frac{cos^3 x}{3} + C = frac{1}{3} cos^3 x — cos x + C $$
    Ответ

    $$ int sin^3 x dx = frac{1}{3} cos^3 x — cos x + C $$
    Пример 4
    Вычислить определенный интеграл от синуса: $$ int_0^pi sin x dx $$
    Решение

    Вычисление начнем как в случае с неопределенным интегралом и в конце используем формулу Ньютона-Лейбница $ int_a^b f(x) dx = F(x) bigg |_a^b = F(b)-F(a) $:

    $$ int_0^pi sin x dx = -cos x bigg |_0^pi = -cos pi + cos 0 = -(-1) + 1 = 1+1=2 $$
    Ответ
    $$ int_0^pi sin x dx = 2 $$
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